Download Pulleys - OpenStax CNX

OpenStax-CNX module: m14060
1
Pulleys
∗
Sunil Kumar Singh
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 2.0
†
Abstract
The basic consideration in making useful mechanical arrangements are two folds (i) improve the
convenience of applying force and (ii) reduce the magnitude of force required.
A pulley is a part of a convenient arrangement that makes it possible to transfer force with change of
direction.
Unless otherwise stated, a pulley is considered to have negligible mass and friction.
This is a
relative approximation with respect to mass and friction involved with other elements. Pulleys are used in
dierent combination with other elements almost always with strings and blocks.
It is relatively dicult to fetch a bucket of water from a well with a string as compared to a pulley and
string system. The basic consideration in making useful mechanical arrangements are two folds (i) improve
the convenience of applying force and (ii) reduce the magnitude of force. The example of fetching of a bucket
of water with pulley and string achieves the goal of improving convenience as we nd it easier applying
force in the level of arms horizontally rather than applying force vertically.
Had it been possible to reduce force for doing a mechanical activity, then that would have been wonder
and of course against the well founded tenets of physical laws. What is meant here by reducing force is that
we can fulll a task (which comprises of force and motion) by reducing force at the expense of extending
motion.
The important characterizing aspects of pulley are discussed in the sections named :
•
•
•
Static or xed pulley
Moving pulley
Combination or Multiple pulley system
1 Static pulley
The pulley is xed to a frame. In this situation, we are only concerned with the accelerations of the bodies
connected to the string that passes over the pulley. Since string is a single piece inextensible element, the
accelerations of the bodies attached to it are same.
We are at liberty to choose the direction of acceleration of the blocks attached to the pulley. A wrong
choice will be revealed by the sign of acceleration that we get after solving equations. However, it is a straight
forward choice here as it is obvious that the bigger mass will pull the blocks - string system down.
Example 1
Problem :
Two blocks of masses 10 kg and 20 kg are connected by a string that passes over a
pulley as shown in the gure. Neglecting friction between surfaces, nd acceleration of the blocks
and tension in the string (consider
∗ Version
g = 10 m / s2
1.20: Dec 4, 2008 10:17 am -0600
† http://creativecommons.org/licenses/by/2.0/
http://cnx.org/content/m14060/1.20/
).
OpenStax-CNX module: m14060
2
Static pulley system
Figure 1
Solution :
The blocks are connected by a taut string. Hence, their accelerations are same. Let
us assume directions of accelerations as shown in the gure. Also, let the magnitude of accelerations
be a.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
3
Static pulley system
Figure 2
Free body diagram of body of mass 10 kg
The external forces are (i) weight of block, 10g, and (ii) tension, T, in the string.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
4
Free body diagram
Figure 3
P
Fy = T − 10g = 10a
⇒ T − 10 x 10 = 10a
⇒ T = 100 + 10a
Free body diagram of body of mass 20 kg
The external forces are (i) weight of block, 20g, and (ii) tension, T, in the string.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
5
Free body diagram
Figure 4
P
Fy = 20g − T = 20a
⇒ 20 x 10 − T = 20a
⇒ T = 200 − 20a
From two equations, we have :
⇒ 200 − 20a = 100 + 10a
⇒ a =
100
30
= 3.33 m / s2
Putting value of "a" in either of above two equations, we determine tension as :
⇒ T = 100 + 10 X 3.3 = 133 N
There is a very useful technique to simplify the solution involving "mass-less" string and pulley. As string
has no mass, the motion of the block-string system can be considered to be the motion of a system comprising
of two blocks, which are pulled down by a net force in the direction of acceleration.
Let us consider two blocks of mass "
Let us also consider that
m2 > m1
m1
" and "
m2
" connected by a string as in the previous example.
so that block of mass "
m2
" is pulled down and block of mass "
pulled up. Let "a" be the acceleration of the two block system.
Now the force pulling the system in the direction of acceleration is :
F = m2 g − m1 g = (m2 − m1 ) g
http://cnx.org/content/m14060/1.20/
m1
" is
OpenStax-CNX module: m14060
6
The total mass of the system is :
m = m1 + m2
Applying law of motion, the acceleration of the system is :
a=
(m2 − m1 ) g
F
=
m
(m1 + m2 )
(1)
Clearly, this method to nd acceleration is valid when the block - string system can be combined i.e.
accelerations of the constituents of the system are same. We can check the ecacy of this technique, using
the data of previous example. Here,
m1 = 10 kg;
m2 = 20 kg
The acceleration of the block is :
a=
(20 − 10) X10
(m2 − m1 ) g
=
(m1 + m2 )
(10 + 20)
a=
10
= 3.33 m/s2
3
2 Moving pulley
Moving pulley diers to static pulley in one important respect.
The displacements of pulley and block,
which is attached to the string passing over it, may not be same. As such, we need to verify this aspect
while applying force law. The point is brought out with clarity in the illustration explained here. Here, we
consider a block attached to a string, which passes over a mass-less pulley. The string is xed at one end
and the Pulley is pulled by a force in horizontal direction as shown in the gure.
Moving pulley system
Figure 5
In order to understand the relation of displacements, we visualize that pulley has moved a distance x
to its right. The new positions of pulley and block are as shown in the gure. To analyze the situation, we
use the fact that the length of string remains same in two situations. Now,
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
7
Moving pulley system
Figure 6
Length of the string, L, in two situations are given as :
L = AB + BC = AB + BD + ED
⇒ BC = BD + ED
⇒ ED = BC − BD
Now, displacement of the block is :
⇒ CE = CD − ED = CD − BC + BD = BD + BD = 2BD = 2x
Note that for every displacement x of pulley, the displacement of block is 2x. We can appreciate this
fact pictorially as shown in the gure below :
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
8
Moving pulley system
Figure 7
Further, as acceleration is second derivative of displacement with respect to time, the relation between
acceleration of the block (
aB
) and pulley (
aP
) is :
⇒ aB = 2aP
This is an important result that needs some explanation.
(2)
It had always been emphasized that the
acceleration of a taut string is always same through out its body. Each point of a string is expected to have
same velocity and acceleration! What happened here ? One end is xed, while other end is moving with
acceleration. There is, in fact, no anomaly. Simply, the acceleration of the pulley is also reected in the
motion of the loose end of the string as they are in contact and that the motion of the string is aected by
the motion of the pulley.
But the point is made. The accelerations of two ends of a string need not be same, when in contact with
a moving body with a free end.
Example 2
Problem :
A block of mass, m is connected to a string, which passes over a smooth pulley as
shown in the gure. If a force F acts in horizontal direction, nd the accelerations of the pulley
and block.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
9
Moving pulley system
Figure 8
Solution :
Let us consider that the acceleration of pulley is a in the direction of applied
force. Now as analyzed before, acceleration of the block is 2a and is in the same direction as that
of pulley.
Moving pulley system
Figure 9
As motion is conned to x-direction, we draw free body diagram considering forces in x-direction
only.
Free body diagram of pulley
The external forces are (i) force, F, (ii) tension, T, in the string and (ii) tension, T, in the string.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
10
Free body diagram
Figure 10
P
Fx = F − 2T = mp ap
As mass of the pulley is zero,
⇒ F = 2T
⇒ T =
This is an unexpected result.
F
2
The pulley is actually accelerated, but the forces on it form a
balanced force system. This apparent contradiction of force law is due to our approximation that
pulley is mass-less.
Free body diagram of block
The external force is (i) tension, T, in the string.
Free body diagram
Figure 11
P
Fx = T = max = 2ma
Combining two equations, we have :
⇒ a =
http://cnx.org/content/m14060/1.20/
F
4m
OpenStax-CNX module: m14060
11
Thus, acceleration of the pulley is F/4m and that of block is F/2m.
3 Combination or Multiple pulley system
Multiple pulleys may involve combination of both static and moving pulleys. This may involve combining
characterizing aspects of two systems.
Let us consider one such system as shown in the gure.
Multiple pulley system
Figure 12
Let us consider that accelerations of the blocks are as shown in the gure. It is important to note that we
have the freedom to designate direction of acceleration without referring to any other consideration. Here,
we consider all accelerations in downward direction.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
12
Multiple pulley system
Figure 13
We observe that for given masses, there are ve unknowns
a1 , a2 , a3 , T1
and
T2
.
Whereas the
free body diagram corresponding to three blocks provides only three independent relations. Thus, all ve
unknowns can not be evaluated using three equations.
However, we can add two additional equations; one that relates three accelerations and the one that
relates tensions in the two strings. Ultimately, we get ve equations for ve unknown quantities.
1: Accelerations
The pulley A is static. The accelerations of block 1 and pulley B are, therefore, same. The pulley
B, however, is moving. Therefore, the accelerations of blocks 2 and 3 may not be same as discussed for the
case of static pulley. The accelerations of blocks 2 and 3 with respect to moving pulley are dierent than
their accelerations with respect to ground reference.
We need to know the relation among accelerations of block 1, 2 and 3 with respect ground.
In order
to obtain this relation, we rst establish the relation among the positions of moving blocks and pulley
with respect to some xed reference.
Then we can obtain relation among accelerations by taking second
dierentiation of position with respect to time.
Important to understand here is that these positions are
measured with respect to a xed reference. As such, their dierentiation will yield accelerations of the blocks
with respect to xed reference i.e. ground reference. Let the positions be determined from a horizontal datum
drawn through the static pulley as shown in the gure.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
13
Multiple pulley system
Figure 14
Now, the lengths of the two strings are constant. Let they be
L1
and
L2
.
L1 = x0 + x1
L2 = ( x2 − x0 ) + ( x3 − x0 ) = x2 + x3 − 2x0
Eliminating
x0
, we have :
⇒ L2 = x2 + x3 − 2L1 + 2x1
⇒ x2 + x3 + 2x1 = 2L1 + L2 =
constant
This is the needed relation for the positions. We know that acceleration is second derivative of position
with respect to time. Hence,
⇒ a2 + a3 + 2a1 = 0
(3)
This gives the relation of accelerations involved in the pulley system.
2: Tensions in the strings
We can, now, nd the relation between tensions in two strings by considering the free body diagram of
pulley B as shown in the gure.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
14
Free body diagram
Figure 15
Here,
T1 = 2T2
note: This result appears to be simple and on expected line. But it is not so. Note that pulley
"B" itself is accelerated.
The result, on the other hand, is exactly same as for a balanced force
system. In fact this equality of forces in opposite direction is possible, because we have considered
that pulley has negligible mass.
This aspect has been demonstrated in the force analysis of the
example given earlier (you may go through the example again if you have missed the point).
3: Free body diagrams of the blocks
The free body diagrams of the blocks are as shown in the gure.
http://cnx.org/content/m14060/1.20/
(4)
OpenStax-CNX module: m14060
15
Free body diagram
Figure 16
m1 g − T1 = m1 a1
m2 g − T2 = m2 a2
(5)
m3 g − T2 = m3 a3
Thus, we have altogether 5 equations for 5 unknowns.
There is one important aspect of the motions of blocks of mass "
m2
" and "
m3
" with respect to moving
pulley "B". The motion of blocks take place with respect to an accelerating pulley. Thus, interpretation
of the acceleration must be specic about the reference (ground or moving pulley). We should ensure that
all measurements are in the same frame. In the methods, described above we have considered accelerations
with respect to ground. Thus, if acceleration is given with respect to the moving pulley in an analysis, then
we must rst change value with respect to the ground.
Example 3
Problem :
In the arrangement shown in the gure, mass of A = 5 kg and mass of B = 10 kg.
Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Find
the accelerations of blocks.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
16
Pulley block arrangement
Figure 17:
Pulley block arrangement
Solution :
Let the acceleration of block A be a in the downward direction. Let the tensions
in the string be T1 and T2 . See gure below showing forces acting on the blocks and moving
pulley. The force analysis of the block A yields :
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
17
Pulley block arrangement
Figure 18:
Pulley block arrangement
⇒ 5g − T1 = 5a
⇒ 50 − T1 = 5a
From constraint of string length, we see that :
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
18
Pulley block arrangement
Figure 19:
Pulley block arrangement
x1 − x0 + x2 − x0 + x2 = L
⇒ x1 + 2x2 = L + 2x0 = Constant
Dierentiating twice with respect to time, we get relation between accelerations of two blocks
as :
⇒ a1 = −2a2
⇒ a2 = −
a
a1
=−
2
2
Force analysis of the moving mass-less pulley yields :
⇒ T2 = 2T1
Force analysis of the block B results (refer to the force diagram shown in the beginning) :
⇒ 2T1 − 10X10 = 10X
a
= 5a
2
⇒ 2T1 − 100 = 5a
Simultaneous equations of forces on blocks A and B are :
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
19
⇒ 50 − T1 = 5a
⇒ 2T1 − 100 = 5a
Solving for a, we have :
⇒a=0
Thus, accelerations of two blocks are zero. Note, however, that tensions in the strings are not
zero.
4 Exercises
Exercise 1
(Solution on p. 25.)
In the arrangement shown, the acceleration of block A is 1
m/s2
. Consider string and pulley to
be mass-less and friction absent everywhere in the arrangement. Then,
Pulley block arrangement
Figure 20:
Pulley block arrangement
m/s2 .
2
is 2 m/s .
2
is 0.5 m/s .
2
is 0.75 m/s
(a) The acceleration of block B is 1
(b) The acceleration of block B
(c) The acceleration of block B
(d) The acceleration of block B
.
Exercise 2
Given : mass of block A = 0.5 kg and acceleration of pulley = 1
(Solution on p. 25.)
m/s2
. Consider string and
pulley to be mass-less and friction absent everywhere in the arrangement. Then,
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
20
Pulley block arrangement
Figure 21:
Pulley block arrangement
(a) The force F is 1 N.
(b) The force F is 2 N.
(c) The tension in the string is 1 N.
(d) The tension in the string is 0.5 N.
Exercise 3
(Solution on p. 25.)
In the arrangement shown in the gure, accelerations of blocks A,B and C are
respectively.
aA
,
aB
and
aC
Consider string and pulley to be mass-less and friction absent everywhere in the
arrangement. Then
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
21
Blocks and pulleys arrangement
Figure 22:
Blocks and pulleys arrangement
(a) 2aA + 2aB + aC = 0
(b) aA + aB + 2aC = 0
(c) aA + aB + aC = 0
(d) 2aA + aB + aC = 0
Exercise 4
(Solution on p. 26.)
Given : mass of block A = 10 kg, spring constant = 200 N/m, extension in the string = 0.1
m. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement.
Then,
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
22
Pulley block arrangement
Figure 23:
Pulley block arrangement
(a) The force F is 40 N.
m/s2
m/s2
(b) The acceleration of block A is 4
(c) The acceleration of pulley is 1
.
(d)The tension in the string is 10 N.
Exercise 5
(Solution on p. 27.)
In the arrangement shown, the mass of A and B are 1 kg and 2 kg respectively. Consider string
and pulley to be mass-less and friction absent everywhere in the arrangement. Then,
Pulley block arrangement
Figure 24:
Pulley block arrangement
(a) The tension in the string connected to A is 1 N.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
23
(b) The tension in the string connected to A is 20/3 N.
(c) The tension in the string connected to B is 2 N.
(d) The tension in the string connected to B is 20/3 N.
Exercise 6
(Solution on p. 28.)
Two blocks of mass m and nm are hanging over a pulley as shown in the gure.
Consider
string and pulley to be mass-less and friction absent everywhere in the arrangement. If n>1 and
acceleration of the blocks is g/3, then value of n is :
Pulley block arrangement
Figure 25:
Pulley block arrangement
(a)
Exercise 7
3
2
(b)
5
3
(c) 3
(d) 2
(Solution on p. 29.)
Given : mass of block A = 0.5 kg, mass of block B=1 kg and F = 2.5 N. Consider string and
pulley to be mass-less and friction absent everywhere in the arrangement. Then,
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
24
Pulley block arrangement
Figure 26:
Pulley block arrangement
m/s2 .
2
1 m/s .
(a) The acceleration of block B is 1
(b) The acceleration of block A is
(c) The tension in the string connecting block B and pulley is 1 N.
(d) The tension in the string connected to block A is 2 N.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
25
Solutions to Exercises in this Module
Solution to Exercise (p. 19)
We are required to determine acceleration of block "B" to answer this question. By inspection, we observe
that if block A moves x distance towards right, then the string length is distributed in two equal halves
about hanging pulley. Clearly, if A moves by x, then B moves by x/2. As such, acceleration of A is
twice that of B. As it is given that acceleration of A is 1
m/s2 ,
the acceleration of B is 2
m/s2 .
Hence, option (b) is correct.
Solution to Exercise (p. 19)
Let the acceleration of block A is a. We know that the acceleration of block A is two times that of
pulley. Hence,
a = 1X2 = 2 m/s2
Let the tension in the string be T. Considering forces on block A, we have :
Pulley block arrangement
Figure 27:
Pulley block arrangement
T = ma = 0.5X2 = 1 N
Considering forces on the mass-less pulley, we have :
⇒ F = 2T = 2X1 = 2 N
Hence, options (b) and (c) are correct.
Solution to Exercise (p. 20)
Here, we need to determine relation of accelerations of blocks using constraint of string length.
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
26
Blocks and pulleys arrangement
Figure 28:
Blocks and pulleys arrangement
xA + xA − x0 + xB − x0 + xB − x0 + xC − x0 = L
⇒ 2xA + 2xB + xC = L + 3x0 = constant
Dierentiating two times, we have :
⇒ 2aA + 2aB + aC = 0
Hence, option (a) is correct.
Solution to Exercise (p. 21)
Let T be the tension in the string. Here, spring force is equal to tension in the string.
T = kx = 200X0.1 = 20 N
Let acceleration of block A is a. Considering forces on block A, we have :
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
27
Pulley block arrangement
Figure 29:
Pulley block arrangement
⇒ T = ma = 10Xa
⇒a=
20
T
=
= 2 m/s2
10
10
Considering forces on the mass-less pulley, we have :
⇒ F = 2T = 2X20 = 40
N
Now, acceleration of pulley is half of the acceleration of block. Hence, acceleration of pulley is 1
m/s2
.
Hence, options (a) and (c) are correct.
Solution to Exercise (p. 22)
Let the tensions in the strings connected to blocks A and B be T1
and T2
respectively. We see
here that only force making block A to move on the table is tension in the string connected to it.
Let
acceleration of block A is a. From constraint relation, we know that the acceleration of A is twice the
acceleration of B. Thus, acceleration of block B is a/2. Applying law of motion for block A, we have :
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
28
Pulley block arrangement
Figure 30:
Pulley block arrangement
⇒ T1 = m1 Xa = 1Xa = a
Consideration of mass-less pulley,
⇒ T2 = 2T1 = 2Xa = 2a
Applying law of motion for block A, we have :
⇒ m2 g − T2 = m2 X
⇒ 2X10 − 2a = 2X
⇒a=
a
2
a
=a
2
20
m/s2
3
Putting this value in the expressions of tensions, we have :
⇒ T1 = a =
20
N
3
⇒ T2 = 2a =
40
N
3
Hence, options (b) is correct.
Solution to Exercise (p. 23)
We consider blocks and string as one system. The net external force on the system is
F = nmg − mg = (n − 1) mg
Total mass of the system, M, is :
M = nm + m = (n + 1) m
The acceleration of the system i.e. the acceleration of either block is :
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
29
⇒a=
(n − 1) mg
(n − 1) g
=
(n + 1) m
(n + 1)
According to question,
⇒a=
(n − 1) g
g
=
(n + 1)
3
⇒ 3n − 3 = n + 1
⇒ 2n = 4
⇒n=2
Hence, option (d) is correct.
Solution to Exercise (p. 23)
Let the acceleration of block B is a. Considering forces on block of mass B, we have :
Pulley block arrangement
Figure 31:
Pulley block arrangement
F − T1 = M a
⇒ 2.5 − T1 = 1Xa = a
Considering free body diagram of mass-less pulley,
⇒ T1 = 2T2
The pulley and the block are connected by inextensible string.
As such, their accelerations are same.
Further, we know by constraint analysis that acceleration of block is twice that of pulley. Hence, acceleration
of block B is 2a. Considering forces on block A, we have :
⇒ T2 = mX2a = 0.5X2a = a
http://cnx.org/content/m14060/1.20/
OpenStax-CNX module: m14060
30
Thus,
⇒ T1 = 2Xa = 2a
Putting this value in the force equation of block A,
2.5 − 2a = 0.5a
⇒ a = 1 m/s2
Clearly,
⇒ T1 = 2a = 2X1 = 2 N
⇒ T2 = T1 /2 = 1 N
Hence, option (a) is correct.
http://cnx.org/content/m14060/1.20/